In his 'Ordinary Differential Equations' (sec. 1.2) V.I. Arnold says
"... every pair of curves in the square joining different pairs of
opposite corners must intersect".

This is obvious geometrically but I was wondering how one could go
about proving this rigorously. I have thought of a proof using
Brouwer's Fixed Point Theorem which I describe below. I would greatly
appreciate the group's
comments on whether this proof is right and if a simpler proof is
possible.

We take a square with side of length 1. Let the two curves be
$(x_1(t),y_1(t))$ and $(x_2(t),y_2(t))$ where the $x_i$ and $y_i$ are
continuous functions from $[0,1]$ to $[0,1]$. The condition that the
curves join different pairs of opposite corners implies,
$$(x_1(0),y_1(0)) = (0,0)$$
$$(x_2(0),y_2(0)) = (1,0)$$
$$(x_1(1),y_1(1)) = (1,1)$$
$$(x_2(1),y_2(1)) = (0,1)$$

The two curves will intersect if there are numbers $a$ and $b$ in $[0,1]$
such that

Then $(f,g)$ is a continuous function from $[0,1]\times [0,1]$ into itself and
hence must have a fixed point by Brouwer's Fixed Point Theorem. But at
a fixed point of $(f,g)$ it must be the case that $p(a,b)=0$ and $q(a,b)=0$
so the two curves intersect.

Figuring out what $f$ and $g$ to use and checking the conditions in the
last para is a tedious. Can there be a simpler proof?

I am looking for any rigorous proof, the more elementary the better.
–
Jyotirmoy BhattacharyaAug 13 '10 at 18:23

2

See John Stillwell's answer below, which explains how you can reduce to the case where the curves are piecewise linear, which certainly makes things simpler!
–
EmertonAug 14 '10 at 3:52

1

I like the proofs based on $K_5$ and the polygonal Jordan curve theorem, but if all of them are unwound, the Brouwer fixed point theorem proof is the most direct and transparent.
–
Victor ProtsakAug 14 '10 at 6:27

10 Answers
10

Since the full Jordan curve theorem is quite subtle, it might be worth
pointing out that theorem in question reduces to the Jordan curve theorem
for polygons, which is easier.

Suppose on the contrary that the curves $A,B$ joining opposite corners do not meet.
Since $A,B$ are closed sets, their minimum distance apart is some
$\varepsilon>0$. By compactness, each of $A,B$ can be partitioned into finitely
many arcs, each of which lies in a disk of diameter $<\varepsilon/3$. Then, by
a homotopy inside each disk we can replace $A,B$ by polygonal paths $A',B'$ that join the
opposite corners of the square and are still disjoint.

Also, we can replace $A',B'$ by simple polygonal paths $A'',B''$ by omitting loops.
Now we can close $A''$ to a polygon, and
$B''$ goes from its "inside" to "outside" without meeting it, contrary to the
Jordan curve theorem for polygons.

This is gorgeous! It's possibly worth pointing out quite how elementary the Jordan curve theorem for polygons is, to show how little is being black-boxed here. Fix some point $x_0$ not on $C$, and for any (other) point $x$ not on $C$, look at the line segment from $x$ to $x_0$; count the parity of how many times it crosses $C$ (counting double/none if it hits vertices of $C$). This is well-defined and locally constant on $\mathbb{R}\setminus C$ (this is where we use that $C$ is a polygon); so as a locally constant, surjective function to $\{0,1\}$, it disconnects $\mathbb{R}^2$.
–
Peter LeFanu LumsdaineAug 14 '10 at 6:47

Peter: That proof goes through even when the polygon isn't simple, e.g. a polygonal figure-eight, where the parity function disconnects the plane into more than two components. So, there's a bit more work to do.
–
Per VognsenAug 14 '10 at 14:49

1

@Pierre: thanks, yes, I did mean $\mathbb{R}^2$; and while the segment can't cross $C$ infinitely often (a polygon has finitely many edges by definition, at least in the conventions I know) it could contain an edge of $C$ as a subsegment, in which case we do have to look at what directions the adjacent edges of $C$ go off in. @Per: you're right, of course; this doesn't establish the full Jordan curve theorem; I was just thinking of what was necessary for the application at hand (which just needs disconnectedness plus the fact that the second curve's endpoints are in opposite components).
–
Peter LeFanu LumsdaineAug 14 '10 at 17:34

ORIGINAL: This follows from the fact that the complete graph $K_5$ on five vertices cannot be imbedded in $\mathbb S^2, $ in itself an application of Jordan Curve. If your two square curvy diagonals stay inside the square without intersecting, a fifth point outside the square can be joined to the four vertices by disjoint arcs, thus creating a complete graph on five vertices. Very nice book by James Munkres, "Topology: a first course" where, on page 386 exercise 5, he does the graph on five vertices. Note that the concept of inside for the square uses elementary ideas such as convexity.

EDIT: As mentioned by Henry Wilton in comment below, there may be other routes here. In particular, I have a book by Robin J. Wilson just called Introduction to Graph Theory, second edition, and in section 13, pages 64-67, in which he develops Euler's formula for planar graphs and as a quick corollary shows that $K_5$ and $K_{3,3}$ are nonplanar, these being Theorem 13A and Corollary 13E. It is anybody's guess whether JCT is used implicitly in defining "faces" properly for Euler's formula. I don't know.

Henry, I don't think I know how to sort out first principles here. The Munkres book, same page, in exercise 4, has the reader use JCT to show that $K{3,3}$ is non-planar. I'm going to guess that the three facts are roughly equivalent in the sense of quick proofs in both directions for any pair. So the questions become, does the nonplanarity of the complete graph on five vertices imply JCT quickly, and is there some trickery where each nonplanar graph gives the other, all "quickly."
–
Will JagyAug 14 '10 at 2:45

1

Just to clarify slightly, Thomason quickly observes that if a graph is planar then it has a polygonal embedding in the plane; so you can talk about faces before you've proved the JCT.
–
HJRWAug 14 '10 at 5:07

Edit: Actually the reference shows that the Game of Hex always has a winner => Brouwer Fixed Point theorem => a pair of curves in the square joining opposite corners must intersect. So it does use Brouwer's fixed point theorem, but gives an elementary proof of it.

As much as I like the connection with Hex, Gale comments on, but doesn't give the proof of, the additional statement "not both" in the Hex theorem, which is the Hex analogue of the intersection property. Thus for the purposes of this question, Hex is a distraction.
–
Victor ProtsakAug 14 '10 at 6:09

Yes, at first I thought that they proved it directly from the game of Hex. The result asked for would follow from the result that you can't have a state in which both players have a winning line, but the references only show directly that at least one person must win. However, the first link does have a short proof of the result asked for, albeit using the fixed point theorem (and a proof of that). So, it's not as direct and elementary as I thought at first. Still, it answers the question.
–
George LowtherAug 14 '10 at 12:19

The degree of the usual diagonals intersecting each other is 1. One at a time, deform the diagonals via straight-line homotopies to the desired curves. This should preserve Brouwer degree. Lastly, Brouwer degree is well-defined even for continuous functions (using a smooth approximation), and non-zero Brouwer degree implies an intersection.

Alternately, you could artificially avoid the phrase "Brouwer degree" and directly track what happens to the intersection point under the homotopy.

I'm not sure that the last paragraph avoiding the degrees works: the curves may intersect in more than one point, so it's challenging to "directly track what happens to the intersection point".
–
Victor ProtsakAug 14 '10 at 6:19

A homological proof would use the intersection form of the torus. if you consider these paths as based loops on the torus, you see that they are represented as (1,1), and (1,-1), in terms of the standard homology generators. knowing that the intersection form is (0 1; -1 0), we find that the intersection index

Q(v,w) = (1,1)(0 1; -1 0)(1,-1)^t = 2

they already intersect once at the origin, so they must intersect somewhere else in the square. However, you must already have had to compute the intersection form.

How about the following, using the Nested Intervals Theorem (which was in my 2nd year Calculus text) which says the intersection of a nested sequence of closed intervals in $\mathbb{R}$ is non-empty. Here goes the proof:

We construct recursively a nested sequence $I_j := [a_j, b_j]$ of closed intervals for $j \geq 0$. Let $I_0 := [0,1]$. For every $j \geq 0$, construct $I_{j+1}$ as follows: let $m_j$ be the midpoint of $I_j$. If the curves intersect at $t = m_j$, then we are done, so stop the sequence. Otherwise set $I_{j+1}$ to be $[a_j, m_j]$ or $[m_j, b_j]$ depending on whether the curves "switch from left to right" on the first sub-interval or the 2nd (let's say you always make sure that $c_1$ is to the "left" of $c_2$ at $t = a_j$ and to the "right" of $c_2$ at $t = b_j$).

If the sequence is finite, then the curves must intersect, as noted above. So assume the sequence is infinite. The Nested Intervals Theorem and the fact that the length decreases by a factor of 2 at every step implies that $\cap_{j=0}^\infty I_j = \lbrace t\rbrace$ for some $t \in [0,1]$. Then we must have $c_1(t) = c_2(t)$.

I seems that your proof is valid for funtions graphs and no generalized curves: when write "switch from left to right" I guess that you to consider the intersetion of the two curves by the line parallel to a cartesian axis $t$: $t=m_j$, and you suppose to have only two intesection (one for curve) . But this isn't true for general curves. ANyway your proof is adaptable: curves are compact and intesection by a interval is a closed, you have anyway a first point at lest and a first point at right..the rest is the some.
–
Buschi SergioJul 26 '13 at 8:46

i.e. the two boundary components of $I \times I$, as paths from $(0,0) \to (1,1)$.

Then we see that $f(a_1(t))$ is a path that starts at the north pole of the circle, and ends at the south pole, and traverses clockwise, whereas $f(a_2(t))$ starts and ends the same, but traverses counterclockwise. Now $f(I \times I)$ provides a homotopy between these paths. However, they are not homotopic as paths in the circle. This gives a contradiction, and hence the paths must intersect.

Isn't a path in $S^1$ contractible to a point?
–
Piero D'AnconaSep 2 '10 at 12:24

By 'paths', i mean paths starting at the top of the circle, and ending at the bottom.
–
Ryan MicklerSep 2 '10 at 13:29

Perhaps i should have been more clear. f(IxI) is a homotopy 'of paths that start at the top of the circle and end at the bottom'.
–
Ryan MicklerSep 2 '10 at 13:34

1

I guess a more elegant way to see it, is to consider again, the map f: I x I -> S^1, when restricted to the boundary, d(IxI) ~= S^1, we find (from the argument above) that f_d(IxI) : S^1 -> S^1 winds once, but f_IxI gives a homotopy from this map to the constant map, hence a contradiction.
–
Ryan MicklerSep 2 '10 at 14:01

1

After thinking about it some more. This question is really a generalization of the intermediate value theory. The IVT is really a homotopy theory question, where you are detecting pi_0(R-{0}), in this case, we are detecting pi_1(R^2-{0}).
–
Ryan MicklerSep 3 '10 at 4:50

This is the main step of the proof that the plane (in this case, the square) has topological dimension 2. You can find a proof (as elementary as I could make it) in my text Measure, Topology, and Fractal Geometry. In particular, no previous knowledge of algebraic topology is required.